What are the zeroes of the polynomial function f(x)=x^3+2x^2-x-2

Mathematics

2 answers:

lakkis [162]3 years ago

7 0

Answer: -2, -1, 1

<u>Step-by-step explanation:</u>

f(x) = x³ + 2x² - x - 2

0 = x³ + 2x² - x - 2

= x²(x + 2) -1(x + 2)

= (x² - 1) (x + 2)

= (x - 1) (x + 1) (x + 2)

0 = x - 1 0 = x + 1 0 = x + 2

1 = x -1 = x -2 = x

MatroZZZ [7]3 years ago

5 0

$f(x)=x^3+2x^2-x-2\\\\f(x)=0\iff x^3+2x^2-x-2=0\\\\x^2(x+2)-1(x+2)=0\\\\(x+2)(x^2-1)=0\iff x+2=0\ \vee\ x^2-1=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2}\\\\x^2-1=0\qquad\text{add 1 to both sides}\\\\x^2=1\to x=\pm\sqrt1\\\\\boxed{x=-1\ \vee\ x=1}\\\\Answer:\ -2,\ -1,\ 1$

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Need help with this question

AleksAgata [21]

First, let's find the slope of the line using the slope formula, which is:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

(( $x_1$ , $y_1$ ) and ( $x_2$ , $y_2$ ) are points on the line)

In context of this problem, we can use the formula to find the slope of the line between the two points:

$m = \dfrac{-2 -6}{7 - (-1)} = \dfrac{-8}{8} = -1$

Now, we can use the slope in the point-slope formula, which will help us find the final equation of the line. (For reference, the point-slope formula is $(y - y_1) = m(x - x_1)$ where ( $x_1$ , $y_1$ ) is a point on the line)